10-327/Classnotes for Thursday November 4

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See some blackboard shots at BBS/10_327-101104-142342.jpg.

Dror's notes above / Student's notes below
  • Question: Regarding the proof of the Lebesque Number Lemma, let T(x) be the function we were working with in the proof. I am confused with how we reached the conclusion that if d(x,y)<E, then T(y)>= T(x) - E. I know that it was said that this is just an application of the triangle inequality, but I am having a bit of trouble seeing that. Hopefully someone can make this point a bit clearer for me. Thanks! Jason.
    • If you could find a ball of radius 7 around x which fits inside some set U, and you move x just a 1 unit away to y, then by the triangle inequality the ball of radius 6 around y is entirely contained inside the ball of radius 7 around x so it is entirely contained in U. Drorbn 18:37, 6 November 2010 (EDT)
      • I have some doubts with Lebesgue number lemma too.. this delta(x) isn't a radius that we can fit a ball inside one of the U's. It is the supremum of all possible radius. Wouldn't that give us a problem? Don't we need to subtract some small positive value and then find a valid radius? I am not sure if there should be some technicality involved here.-Kai

And once we found this delta(x_0) we should divide by 2 so that delta(x)>delta(x_0) for all x in X? Because it is not necessarily true that we can find a ball of radius delta(x_0) around x_0 to fit into ball?-Kai

  • I think you are correct, (if you follow the proof on the blackboard shots to the letter) as an example of what you are talking about take the open cover of [0,1] using sets of the form [0,x) for x<3/4 and (y,1] for y>1/4. Then \Delta (x)\geq 1/4 and at 1/2 \Delta (1/2)=1/4 but no ball of radius 1/4 around 1/2 fits inside any of the sets. So we will need to take a smaller value than inf(\Delta (x)) as our value of \delta_0 for Lebesgue's Lemma. Dividing by two as you suggest should work fine to fix this problem... Maybe it's just me, but in proof's like this I always feel the urge to divide the final answer by 2, just in case I mixed up some some strict inequality, with a non-strict one somewhere - John
    • Thanks John very nice example. Can you also help me with this question? "delta(x) isn't a radius that we can fit a ball inside one of the U's. It is the supremum of all possible radius. Wouldn't that give us a problem in showing delta(y)>=delta(x)-epsilon when d(x,y)<epsilon?"-Kai
      • I don't think there is any problem with this step.

Without loss of generality  \epsilon < \Delta (x) otherwise the condition holds vacuously. Suppose we construct a ball of radius r such that  r > \epsilon around x so that it is a subset of some U in the cover, then we can construct a ball of radius r - \epsilon around y such that this ball is also in U. So  \Delta (y) \geq r-\epsilon for all r such that a ball of radius r exists around x as in the proof. Which implies  \Delta (y) \geq sup(r) - \epsilon implies  \Delta (y) \geq \Delta (x) - \epsilon . Hopefully this makes sense and works - John -Yes. Makes sense. Nice and concise thanks! -Kai